Which unit rate is the lowest price per ounce? (1 point) Choice A: 30 ounces of chocolate chips for $2.49 Choice B: 40 ounces of chocolate chips for $3.32 Question 1 options: 1) Choice A 2) Choice B 3) The unit rates are equal. 4) The unit rates cannot be determined.

Answer:

where are the options

Step-by-step explanation:

Answer:

it cost $8, it was a swimsuit and he bought it

Step-by-step explanation:

Answer:

Hi it’s me

Step-by-step explanation:

7is the first one

10 is the second

Nvm you already got the answer cause yo asked this 2 weeks ago anyways hi

Answer:

0.3495.

Step-by-step explanation:

............................

Answer:

A

Step-by-step explanation:

Answer:

x = -2

Step-by-step explanation:

[tex]\text {Hello! Let's Solve this Problem!}[/tex]

[tex]\text {\underline {The First Step is to Add 4}}[/tex]

[tex]\text {3x-4+4=14+4}[/tex]

[tex]\text {3x=18}[/tex]

[tex]\text {\underline {The Final Step is to Divide 3}}[/tex]

[tex]\text {3x/3=18/3}[/tex]

[tex]\text {Your Answer Would Be:}[/tex]

[tex]\Huge\boxed {x=6}[/tex]

[tex]\text {Best of Luck!}[/tex]

[tex]\huge\text{$x=\boxed{6}$}[/tex]

We can solve for the variable [tex]x[/tex] by isolating it on one side of the equals sign.

Start by adding [tex]4[/tex] to both sides to cancel out the [tex]-4[/tex] on the left side of the equation.

[tex]\begin{aligned}3x-4+4&=14+4\\3x&=18\end{aligned}[/tex]

Now divide both sides of the equation by [tex]3[/tex] since [tex]3x[/tex] is the same as [tex]3*x[/tex] and we want to isolate [tex]x[/tex], keeping in mind that anything multiplied by [tex]1[/tex] is itself.

[tex]\begin{aligned}3x&=18\\3*x&=18\\3*x\div3&=18\div3\\x*3\div3&=18\div3\\x*1&=6\\x&=\boxed{6}\end{aligned}[/tex]

Answer:

Step-by-step explanation:

t22 = a + (n- 1)d

t22 = 3 + 21*4

t22 = 3 + 84

t22 = 87

Answer:

The solution is too long. So, I included them in the explanation

Step-by-step explanation:

This question has missing details. However, I've corrected each question before solving them

Required: Determine the inverse

1:

[tex]f(x) = 25x - 18[/tex]

Replace f(x) with y

[tex]y = 25x - 18[/tex]

Swap y & x

[tex]x = 25y - 18[/tex]

[tex]x + 18 = 25y - 18 + 18[/tex]

[tex]x + 18 = 25y[/tex]

Divide through by 25

[tex]\frac{x + 18}{25} = y[/tex]

[tex]y = \frac{x + 18}{25}[/tex]

Replace y with f'(x)

[tex]f'(x) = \frac{x + 18}{25}[/tex]

2. [tex]g(x) = \frac{12x - 1}{7}[/tex]

Replace g(x) with y

[tex]y = \frac{12x - 1}{7}[/tex]

Swap y & x

[tex]x = \frac{12y - 1}{7}[/tex]

[tex]7x = 12y - 1[/tex]

Add 1 to both sides

[tex]7x +1 = 12y - 1 + 1[/tex]

[tex]7x +1 = 12y[/tex]

Make y the subject

[tex]y = \frac{7x + 1}{12}[/tex]

[tex]g'(x) = \frac{7x + 1}{12}[/tex]

3: [tex]h(x) = -\frac{9x}{4} - \frac{1}{3}[/tex]

Replace h(x) with y

[tex]y = -\frac{9x}{4} - \frac{1}{3}[/tex]

Swap y & x

[tex]x = -\frac{9y}{4} - \frac{1}{3}[/tex]

Add [tex]\frac{1}{3}[/tex] to both sides

[tex]x + \frac{1}{3}= -\frac{9y}{4} - \frac{1}{3} + \frac{1}{3}[/tex]

[tex]x + \frac{1}{3}= -\frac{9y}{4}[/tex]

Multiply through by -4

[tex]-4(x + \frac{1}{3})= -4(-\frac{9y}{4})[/tex]

[tex]-4x - \frac{4}{3}= 9y[/tex]

Divide through by 9

[tex](-4x - \frac{4}{3})/9= y[/tex]

[tex]-4x * \frac{1}{9} - \frac{4}{3} * \frac{1}{9} = y[/tex]

[tex]\frac{-4x}{9} - \frac{4}{27}= y[/tex]

[tex]y = \frac{-4x}{9} - \frac{4}{27}[/tex]

[tex]h'(x) = \frac{-4x}{9} - \frac{4}{27}[/tex]

4:

[tex]f(x) = x^9[/tex]

Replace f(x) with y

[tex]y = x^9[/tex]

Swap y with x

[tex]x = y^9[/tex]

Take 9th root

[tex]x^{\frac{1}{9}} = y[/tex]

[tex]y = x^{\frac{1}{9}}[/tex]

Replace y with f'(x)

[tex]f'(x) = x^{\frac{1}{9}}[/tex]

5:

[tex]f(a) = a^3 + 8[/tex]

Replace f(a) with y

[tex]y = a^3 + 8[/tex]

Swap a with y

[tex]a = y^3 + 8[/tex]

Subtract 8

[tex]a - 8 = y^3 + 8 - 8[/tex]

[tex]a - 8 = y^3[/tex]

Take cube root

[tex]\sqrt[3]{a-8} = y[/tex]

[tex]y = \sqrt[3]{a-8}[/tex]

Replace y with f'(a)

[tex]f'(a) = \sqrt[3]{a-8}[/tex]

6:

[tex]g(a) = a^2 + 8a- 7[/tex]

Replace g(a) with y

[tex]y = a^2 + 8a - 7[/tex]

Swap positions of y and a

[tex]a = y^2 + 8y - 7[/tex]

[tex]y^2 + 8y - 7 - a = 0[/tex]

Solve using quadratic formula:

[tex]y = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}[/tex]

[tex]a = 1[/tex] ; [tex]b = 8[/tex]; [tex]c = -7 - a[/tex]

[tex]y = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}[/tex] becomes

[tex]y = \frac{-8 \±\sqrt{8^2 - 4 * 1 * (-7-a)}}{2 * 1}[/tex]

[tex]y = \frac{-8 \±\sqrt{64 + 28 + 4a}}{2 * 1}[/tex]

[tex]y = \frac{-8 \±\sqrt{92 + 4a}}{2 * 1}[/tex]

[tex]y = \frac{-8 \±\sqrt{92 + 4a}}{2 }[/tex]

Factorize

[tex]y = \frac{-8 \±\sqrt{4(23 + a)}}{2 }[/tex]

[tex]y = \frac{-8 \±2\sqrt{(23 + a)}}{2 }[/tex]

[tex]y = -4 \±\sqrt{(23 + a)}[/tex]

[tex]g'(a) = -4 \±\sqrt{(23 + a)}[/tex]

7:

[tex]f(b) = (b + 6)(b - 2)[/tex]

Replace f(b) with y

[tex]y = (b + 6)(b - 2)[/tex]

Swap y and b

[tex]b = (y + 6)(y - 2)[/tex]

Open Brackets

[tex]b = y^2 + 6y - 2y - 12[/tex]

[tex]b = y^2 + 4y - 12[/tex]

[tex]y^2 + 4y - 12 - b = 0[/tex]

Solve using quadratic formula:

[tex]y = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}[/tex]

[tex]a = 1[/tex] ; [tex]b = 4[/tex]; [tex]c = -12 - b[/tex]

[tex]y = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}[/tex] becomes

[tex]y = \frac{-4\±\sqrt{4^2 - 4 * 1 * (-12-b)}}{2*1}[/tex]

[tex]y = \frac{-4\±\sqrt{4^2 - 4 *(-12-b)}}{2}[/tex]

Factorize:

[tex]y = \frac{-4\±\sqrt{4(4 - (-12-b))}}{2}[/tex]

[tex]y = \frac{-4\±2\sqrt{(4 - (-12-b))}}{2}[/tex]

[tex]y = \frac{-4\±2\sqrt{(4 +12+b)}}{2}[/tex]

[tex]y = \frac{-4\±2\sqrt{16+b}}{2}[/tex]

[tex]y = -2\±\sqrt{16+b}[/tex]

Replace y with f'(b)

[tex]f'(b) = -2\±\sqrt{16+b}[/tex]

8:

[tex]h(x) = \frac{2x+17}{3x+1}[/tex]

Replace h(x) with y

[tex]y = \frac{2x+17}{3x+1}[/tex]

Swap x and y

[tex]x = \frac{2y+17}{3y+1}[/tex]

Cross Multiply

[tex](3y + 1)x = 2y + 17[/tex]

[tex]3yx + x = 2y + 17[/tex]

Subtract x from both sides:

[tex]3yx + x -x= 2y + 17-x[/tex]

[tex]3yx = 2y + 17-x[/tex]

Subtract 2y from both sides

[tex]3yx-2y =17-x[/tex]

Factorize:

[tex]y(3x-2) =17-x[/tex]

Make y the subject

[tex]y = \frac{17 - x}{3x - 2}[/tex]

Replace y with h'(x)

[tex]h'(x) = \frac{17 - x}{3x - 2}[/tex]

9:

[tex]h(c) = \sqrt{2c + 2}[/tex]

Replace h(c) with y

[tex]y = \sqrt{2c + 2}[/tex]

Swap positions of y and c

[tex]c = \sqrt{2y + 2}[/tex]

Square both sides

[tex]c^2 = 2y + 2[/tex]

Subtract 2 from both sides

[tex]c^2 - 2= 2y[/tex]

Make y the subject

[tex]y = \frac{c^2 - 2}{2}[/tex]

[tex]h'(c) = \frac{c^2 - 2}{2}[/tex]

10:

[tex]f(x) = \frac{x + 10}{9x - 1}[/tex]

Replace f(x) with y

[tex]y = \frac{x + 10}{9x - 1}[/tex]

Swap positions of x and y

[tex]x = \frac{y + 10}{9y - 1}[/tex]

Cross Multiply

[tex]x(9y - 1) = y + 10[/tex]

[tex]9xy - x = y + 10[/tex]

Subtract y from both sides

[tex]9xy - y - x = y - y+ 10[/tex]

[tex]9xy - y - x = 10[/tex]

Add x to both sides

[tex]9xy - y - x + x= 10 + x[/tex]

[tex]9xy - y = 10 + x[/tex]

Factorize

[tex]y(9x - 1) = 10 + x[/tex]

Make y the subject

[tex]y = \frac{10 + x}{9x - 1}[/tex]

Replace y with f'(x)

[tex]f'(x) = \frac{10 + x}{9x -1}[/tex]

Answer:

Step-by-step explanation:

Plug in the stated values for x

(a) 2(1)^4 + 3(1)^3 - (1)^3 + 2(1) - 3

   2 + 3  - 1 + 2 - 3 = 1 + 2 = 3

  f(1) = 3

(b) 2(-1)^4 + 3(-1)^3 - (-1)^2 + 2(-1) - 3

2 * 1 + 3(-1) - 1 + 2(-1) - 3

2 + -3 -1 -2 - 3 = 2 - 9 = -7

f(-1) = -7

(c) 2(0) + 3(0) - 0 + 0 - 3

0 - 3 = -3

f(0) = -3

d) 2(-1/2)^4 + 3(-1/2)^3 - 2(-1/2) - 3

1/8 - 3/8 - (-1/2)^2 - 2 * 1/2 - 3

16/4 - 1/4 - 1/4 = -16-1-1/4 = -18/4

f(-1/2) = -9/2

Answer:

have:

[tex]\frac{sin}{1-cot}+\frac{cos}{1-tan}\\\\=\frac{sin}{1-\frac{cos}{sin} }+\frac{cos}{1-\frac{sin}{cos} }\\\\=\frac{sin^{2} }{sin-cos}+\frac{cos^{2} }{cos-sin} \\\\=\frac{sin^{2} }{sin-cos}-\frac{cos^{2} }{cos-sin}\\\\=\frac{sin^{2}-cos^{2} }{sin - cos}\\\\=\frac{(sin-cos)(sin+cos)}{sin-cos}\\\\=sin+cos[/tex]

Step-by-step explanation:

Answer:

The solutions to the system of the equations by the elimination method will be:

[tex]x=2,\:z=-1,\:y=2[/tex]

Step-by-step explanation:

Given the system of the equations

[tex]2x\:+\:2y\:+z\:=\:7[/tex]

[tex]-x-\:y\:+z\:=\:-5[/tex]

[tex]x+3y-4z=12[/tex]

solving the system of the equations by the elimination method

[tex]\begin{bmatrix}2x+2y+z=7\\ -x-y+z=-5\\ x+3y-4z=12\end{bmatrix}[/tex]

[tex]\mathrm{Multiply\:}-x-y+z=-5\mathrm{\:by\:}2\:\mathrm{:}\:\quad \:-2x-2y+2z=-10[/tex]

[tex]\begin{bmatrix}2x+2y+z=7\\ -2x-2y+2z=-10\\ x+3y-4z=12\end{bmatrix}[/tex]

[tex]-2x-2y+2z=-10[/tex]

[tex]+[/tex]

[tex]\underline{2x+2y+z=7}[/tex]

[tex]3z=-3[/tex]

[tex]\begin{bmatrix}2x+2y+z=7\\ 3z=-3\\ x+3y-4z=12\end{bmatrix}[/tex]

[tex]2x+6y-8z=24[/tex]

[tex]-[/tex]

[tex]\underline{2x+2y+z=7}[/tex]

[tex]4y-9z=17[/tex]

[tex]\begin{bmatrix}2x+2y+z=7\\ 3z=-3\\ 4y-9z=17\end{bmatrix}[/tex]

Rearranging the equations

[tex]\begin{bmatrix}2x+2y+z=7\\ 4y-9z=17\\ 3z=-3\end{bmatrix}[/tex]

solve [tex]3z=-3[/tex] for z:

[tex]z=-1[/tex]

[tex]\mathrm{For\:}4y-9z=17\mathrm{\:plug\:in\:}z=-1[/tex]

solve  [tex]4y-9\left(-1\right)=17[/tex] for y:

[tex]4y-9\left(-1\right)=17[/tex]

[tex]4y+9=17[/tex]

[tex]4y=8[/tex]

[tex]y=2[/tex]

[tex]\mathrm{For\:}2x+2y+z=7\mathrm{\:plug\:in\:}z=-1,\:y=2[/tex]

solve [tex]2x+2\cdot \:2-1=7[/tex] for x:

[tex]2x+2\cdot \:2-1=7[/tex]

[tex]2x+3=7[/tex]

[tex]2x=4[/tex]

[tex]x=2[/tex]

Therefore, the solutions to the system of the equations by the elimination method will be:

[tex]x=2,\:z=-1,\:y=2[/tex]

Answer:

x < 1

Step by step explanation:

10 + 4x < 14

4x < 14 -10

4x < 4

x < 1

When you graph it on a number line, draw a line that extends from 1 towards the left side (since x is less that 1) and use unshaded circles.

Answer:

I think its C (I think..im doing the test so ill tell u if its right or wrong)

Step-by-step explanation:

Answer:

A

Step-by-step explanation:

Blake should not include the outlier because it overstates the effect of rainfall on temperature.

No, because intergers don't include fractions.

Answer:

No

Step-by-step explanation:

Integers cannot have any fractions.

If you simplify this to get 8/1 and then 8, 8 is an integer

Answer:

Step-by-step explanation:

Given that :

Dimension of rail road toy trains :

Length = 15 inches

Width = 4 inches

Scale = 2/9

Dimension of car on the long Island rail road :

Let length = x

Length = 2/9 of x = 15 inches

= 0.22222x = 15 inches

LENGTH = 67.5 inches

WIDTH = 2/9 of x = 4 inches

0.22222x = 4 inches

Width = 18

Width = 6 inches

Answer:

Wht u need help on?

Step-by-step explanation:

The difference between consecutive first and second integers is increased by 1 thus the next number 19 - x = 5 → x = 14.

A pattern is a specific arrangement of numbers or any term which will follow a specific code or hint.

For example 2,4,6,8 in this, the difference between any two-term is only 2 and this will be its specific pattern.

As per the given pattern,29, 28, 26, 23, 19

29 - 28 = 1

28 - 26 = 2

26 - 23 = 3

23 - 19 = 4

Since the difference is increasing by 1

So the difference of 19 and the next number (x) will be 4 + 1 = 5.

19 - x = 5

x = 14

Hence "The difference between consecutive first and second integers is increased by 1 thus the next number 19 - x = 5 → x = 14".

For more information about the pattern,

brainly.com/question/14013716

#SPJ2

Answer:

A ( SAS )

Step-by-step explanation:

Answer:

1) The equation that represents the number of slices of bacon to the number he will eat, assuming he eats 24 grams is;

3·x + 6·y = 24

2) The equation in slope and intercept form is;

y = 4 - 0.5·x

Step-by-step explanation:

1) The given parameters are;

The amount of protein Tommy wants to eat for breakfast = 24 grams

The quantity of protein in a slice of bacon = 3 grams

The quantity of protein in an egg = 6 grams

Let x represent the number of slices of bacon he will eat and let y represent the number of eggs he will eat

The equation that represents the number of slices of bacon to the number he will eat, assuming he eats 24 grams is given as follows;

3·x + 6·y = 24

2) The equation in slope and intercept form is given by making y the subject of the above equation as follows;

3·x + 6·y = 24

6·y = 24 - 3·x

y = 24/6 -1/2·x

y = 4 - 0.5·x, which is in the slope and intercept form y = m·x + c

Where;

m = The slope = -0.5

c = The y-intercept = 4

Answer: 600

Step-by-step explanation:

I just took the test

Answer:1222

Step-by-step explanation:

this is why this the answer

Answers:

===================================================

Explanation:

The largest exponent term is what will go first. Then the term with the second largest goes next, and so on. We count the exponents down. The exponents will count down in this case as 5, 3, 1, 0

Note: -9 = -9*1 = -9m^0 and -11m = -11m^1

So that's why standard form is [tex]\frac{3}{5}m^5 + 7m^3 - 11m-9[/tex]

We can write this as (3/5)m^5 + 7m^3 - 11m - 9

------------------------

The leading term is the first term after getting everything in standard form. The coefficient of the leading term is the leading coefficient. So in this case, it would be 3/5.

The degree is the largest exponent. So the degree is 5. This is a quintic polynomial.

Answer:

The most that they can spend in food is $239

Step-by-step explanation:

Their budget allows for less than $720 for both food and rent.

We are told that this budget allows

twice as much money for rent as

for food.

Thus,

Let x be amount of spent on food.

Thus, money spent on rent is 2x

Thus,

2x + x < 720

3x < 720

x < 720/3

x < 240

Since x should be less than 240,it means that the most that they can spend in food is $239

Answer:

B

Step-by-step explanation:

big line= divide

[tex]-2 1/4 to improper = -9/4\\/ -2/3[/tex]

so -9/4 / -2/3 or

B

Hope this helps :D

Answer:

B!

Step-by-step explanation:

if you calculate the original expression given you get [tex]3\frac{3}{8}[/tex] so i just went down the line calculating the rest of the equations and B was the one equivalent to the original expression! hope this helped.

Answer:

Area of ΔDEF is 2.25 times the area of ΔABC.

Step-by-step explanation:

Since, ΔDEF is the enlarged form of ΔABC,

Scale factor = [tex]\frac{\text{Side length of triangle ABC}}{\text{Corresponding side length of image triangle}}[/tex]

                    = [tex]\frac{15}{10}[/tex]

                    = 1.5

Since, ratio of the areas of image and preimage triangles = (Scale factor)²

[tex]\frac{\text{Area of image triangle}}{\text{Area of triangle ABC}}[/tex] = (Scale factor)²

[tex]\frac{\text{Area of image triangle}}{\text{Area of triangle ABC}}=(1.5)^2[/tex]

Area of image ΔDEF = 2.25(Area of ΔABC)

Therefore, area of ΔDEF is 2.25 times the area of ΔABC.